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My name is Ann [436]
3 years ago
14

PLEASE HELPPPP MEEEE NOWWWW!

Mathematics
1 answer:
Ket [755]3 years ago
6 0

Answer:

2, 6, and 7

Step-by-step explanation:

3n is by itself because theres no other numbers that have the letter n.

6m and -11m are like terms bcs they both have the letter m (variable)

-3 and -18 are both like terms bcs they are both just numbers (or constants whatever)

so you would choose options 2, 6, and 7! hope this helps! :)

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Without evaluating each expression, determine which value is the greatest.
alina1380 [7]

Answer:

756−934 is the greatest because the others are more in the negatives  

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
You eat 18 grams of your daily carbohydrates which is 6% of daily needs what is your daily need
adell [148]

Answer:

300 grams

Step-by-step explanation:

To find the total  number of carbohydrates, you need to find what fraction 6 is of 100, so you can do 100/6, but I just did 18* 100/6 right away, so that you can find 18* the fraction that 6 is of 100. The answer that I got is 300 grams. Hope this helps!

7 0
3 years ago
Please help!!! i need 2.31 as a percent
Amanda [17]

Answer:

%231

Step-by-step explanation:

To get a percent, just multiply the decimal by 100 to get the percent.

<u>Example:  Convert 0.5 into a percentage</u>

0.5 * 100

%50

<u>Problem:  Convert 2.31 into a percentage</u>

2.31 * 100

<em>%231</em>

<em />

Answer:  %231

5 0
3 years ago
Consider the following information about travelers on vacation: 40% check work email, 30% use a cell phone to stay connected to
andrew-mc [135]

Answer:

a. 0.4

b. 0.6

c. 0.6493

Step-by-step explanation:

p(checking work email) = p(A) = 0.40

p(staying connected with cell phone) = p(B) = 0.30

p(having laptop) = p(c) = 0.35

p(checking work mail and staying connected with cell phone) = p(AnB) = 0.16

p(neither A,B or C) = p(AuBuC)

= 1-42.8%

= 0.572

p(A|C) = 88% = 0.88

p(C|B) = 70% = 0.7

a. What is the probability that a randomly selected traveler who checks work email also uses a cell phone to stay connected?

p(B|A) = p(AnB)/p(A)

= 0.16/0.4

= 0.4

b. What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected?

p(B|C) = P(C|B)p(B)/p(C)

= 0.7x0.3/0.35

= 0.6

c. If the randomly selected traveler checked work email and brought a laptop, what is the probability that he/she uses a cell phone to stay connected?

p(A|BnC)

= P(BnAnC)/p(AnC)

= p(AnC) = p(A|C).p(C)

= 0.88x0.35

= 0.308

p(AnBnC) = p(AuBuC)-p(a)-p(b)+ p(AnB)+p(AnC)+p(BnC)

p(BnC) = 0.7x0.3

= 0.21

p(AnBnC) = 0.572-0.4-0.3-0.35+0.16+0.308+0.21

= 0.2

p(A|BnC) = 0.2/0.308

= 0.6493

3 0
3 years ago
Function g can be thought of as a scaled version of f(x)=x^2.<br> Write the equation for g(x).
slega [8]

Answer:

5x to the 2nd power

Step-by-step explanation:

3 0
4 years ago
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